(a) If the switch is thrown to the left (connecting the battery), how much time

Question

(a) If the switch is thrown to the left (connecting the battery), how much time elapses before the current reaches 220 mA?

ms

(b) What is the current in the inductor 10.0 s after the switch is closed?

A
(c) Now the switch is quickly thrown from a to b. How much time elapses before the current falls to 160 mA?
ms

A 140 mH inductor and a 4.50 (a) If the switch is thrown to the left (connecting the battery), how much time elapses before the current reaches 220 mA? ms (b) What is the current in the inductor 10.0 s after the switch is closed? A (c) Now the switch is quickly thrown from a to b. How much time elapses before the current falls to 160 mA? ms Omega resistor are connected with a switch to a 6.00 V battery as shown in Figure P32.27.

Explanation / Answer

Switch thrown to left

At steady state inductor will acts like short circuit

Steady state current =Max curent = 6/4.5 = 4/3

time constant fo charging =L/R = 140 X 10^-3 / 4.5 =0.031

I= Imax (1-e^(-t/timeconstant) Imax =4/3

Putting I =220 mA    t = 5.59 miliseconds

Current after 10 seconds

Put 10 seconds in I= Imax (1-e^(-t/timeconstant) Imax =4/3

I = 4/3 =1.33 Ampere

c

Switch put to b side

Current thriugh will not change instatneously

Inductor will discharge

I = Imax e^(-t/time constant)

0.16 = 3/4 e^(-t/0.031)

t=47.8 miliseconds

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